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\begin{document}

{\bf Solution 1.}

We need $m^2\equiv 7\pmod{25}$. We have $3^3=27\equiv
7\pmod{5}$. Then $(3+5x)^3\equiv 27+3.3^2.5x\pmod{25}$, giving
$7\pmod{25}$ iff $20+135x\equiv 0\pmod{25}$, iff $x\equiv 3\pmod{5}$,
whence $18^3\equiv 7\pmod{25}$.  Hence $m=18$ works.

[S; 2 marks]\bigskip

If $|n^3-7|_3<3^{-1}$ then $n^3\equiv 7\pmod{9}$, which has no
solutions. 

[S; 2 marks]\bigskip

If $A+B+C=0$ in a field with a non-archimedean valuation, and 
$|A|\ge|B|\ge|C|$, then we must have $|A|=|B|$.  Indeed we will either
have $A=B=C=0$ or $|A|=|B|>0$. In our case
$|x^3|_2=2^{-3a}$ or $=0$, $|2y^3|_2=2^{-1-3b}$ or $=0$, and 
$|4z^3|_2=2^{-2-3c}$ or $=0$. Hence two of these can be equal only
when they are both zero.  It follows that $x=y=z=0$. 

[S; 4 marks]\bigskip

We have convergence iff $|x^n/n!|_2\rightarrow 0$. There are
infinitely many values of $n$ for which $|n!|_2=2^{1-n}$, so that we
cannot have convergence if $|x|_2\ge 2^{-1}$.  On the other hand, if
$|x|_2\le 2^{-2}$ then $|x^n/n!|_2\le 2^{-2n}.2^{n-1}$, since we are
given that $|n!|_2\ge 2^{-n}$.  Thus $|x^n/n!|_2\rightarrow 0$ if 
$|x|_2\le 2^{-2}$.  Thus it is necessary and sufficient that
$|x|_2\le 2^{-2}$.

[B or N (optional question, sheet 3); 4 marks]\bigskip

Hensel: Let $K$ be a field, complete w.r.t. a non-archimedean
valuation $|\ldots|$, with valuation ring $R$.  Let $f(x)\in R[x]$, and let
$a\in R$ satisfy 
\[|f(a)|<|f'(a)|^2.\]
Then there is a unique $a^*\in R$ with $f(a^*)=0$ and 
\[|a^{*}-a|\le |f(a)|/|f'(a)|.\]

[B; 3 marks]\bigskip

We use Hasse's bound: 
\[|\# E(\mathbb{F}_p)-(p+1)|\le 2\sqrt{p}\]
for
any elliptic curve $E$ over $\mathbb{F}_p$.  If $p$ is a prime of good
reduction for $y^2=x^3+ax+b$ it follows that the congruence $y^2\equiv
x^3+ax+b\pmod{p}$ has at least $p-2\sqrt{p}$
solutions, after allowing for the point at infinity. Since $p\ge 5$
we deduce that there is at least one solution $(u_0,v_0)$ say, 
which is necessarily
non-singular since there is good reduction at $p$. If the partial
derivative in $x$ is not divisible by $p$ we apply Hensel's Lemma to
$f(t)=x^3+ax+b-v_1^2$ for any $v_1\in\Z_p$ with $v_1\equiv
v_0\pmod{p}$.  Thus there is a solution $(u,v_1)$ for each such
$v_1$.  Similarly if the partial derivative in $y$ is not divisible by
$p$ at $(u_0,v_0)$ we may work with $f(t)=t^2-u_1^3-au_1-b$, taking
any $u_1\equiv u_0\pmod{p}$.

[S; 5 marks]\bigskip

In the case of bad reduction $x^3+ax+b$ has a repeated factor modulo $p$.

[B; 1 mark]\bigskip

The above argument goes through if the reduction to $\mathbb{F}_p$ has
at least one smooth point.  However the reduction takes the form
$y^2=(x+\alpha)^2(x+\beta)$. If $\alpha\not=\beta$ then $(-\beta,0)$
is a smooth point, while if $\alpha=\beta$ then $(1-\alpha,1)$ is
smooth. (In the latter case the partial w.r.t $x$ is 3, while the
partial derivative w.r.t $y$ is 2.  These cannot both vanish.)

[N; 4 marks]


\newpage

{\bf Solution 2.}


A formal group is a power series $F(X,Y)\in R[[X,Y]]$ s.t.
\begin{enumerate}
\item[(i)] $F(X,Y)=X+Y+$ terms of degree 2 or higher.
\item[(ii)] $F(X,F(Y,Z))=F(F(X,Y),Z)$.
\item[(iii)] $F(X,Y)=F(Y,X)$.
\end{enumerate}

[B; 3 marks]\bigskip

Construct $Z_n\in TR[[T]]$ of degree $n$ inductively so that 
$F(T,Z_n)$ has all
terms of degree greater than $n$. By (i) we may start with $Z_1=-T$.
Suppose that $F(T,Z_n)=a_{n+1}T^{n+1}+\ldots$.  Then if
$Z_{n+1}=Z_n+cT^{n+1}$ we have a Taylor series expansion in powers of 
$cT^{n+1}$ as
\[F(T,Z_{n+1})=F(T,Z_n)+cT^{n+1}F_Y(T,Z_n)+\mbox{ terms of degree $\ge
  2n+2$},\]
and since \[F_Y(T,Z_n)=1+\mbox{ higher order terms}\]
by (i), we deduce
that $F(T,Z_{n+1})=F(T,Z_n)+cT^{n+1}+\mbox{ higher order terms}$.  It follows
that $Z_{n+1}$ has the required properties iff $c=-a_{n+1}$.  Finally
$i(T)$ has the required property iff it is the limit of such a
sequence of polynomials $Z_n$, which we have seen can be chosen in
exactly one way.

[B; 3 marks]\bigskip

It follows in the same way that there is a unique $j(T)\in TR[[T]]$
for which $F(i(T),j(T))=0$.  We claim that both $j(T)=T$ and
$j(T)=F(T,0)$ have this property, which will prove that $T=F(T,0)$.
By (iii) we have $F(i(T),T)=F(T,i(T))=0$ so that $j(T)=T$ is
acceptable.  By (ii) $F(i(T),F(T,0))=F(F(i(T),T),0)=F(0,0)=0$, by (iii)
and (i), so $j(T)=F(T,0)$ is also acceptable, as claimed.

[B; 4 marks]\bigskip

Suppose $(x,y)$ is a non-trivial torsion point lying in the kernel of
reduction, so that $|x|_p,|y|_p>1$ and hence $|y|_p=|x|_p^{3/2}$. 
Let $z=-x/y, w=-1/y$, whence $|z|_p,|w|_p<1$.  Then $z$ will be a
torsion element for the (formal) group $F_E(p\Z_p)$ associated to $E$.
It is a fact from the general theory of formal groups that such a
torsion element must have an order which is a power of $p$, say
$p^n$.  Moreover it is also a general fact that one then has 
\[|z|_p\ge |p|_p^{1/(p^n-p^{n-1})}.\]
Except in the case $p^n=2$ we have $1/(p^n-p^{n-1})<1$, which would
give $1>|z|_p>p^{-1}$, a contradiction.  Thus the only possibility is that
$(x,y)$ has order 2 on $E$.  However in this case we will have $y=0$,
contradicting the inequality $|y_p|>1$.

[B; 8 marks]\bigskip


If $x$ or $y$ were not in $\Z$ we would have $|x|_p>1$ or $|y|_p>1$
(and hence both) for some $p$. Then $(x,y)$ would be in the kernel of
reduction, contrary to the above result.

[B; 2 marks]\bigskip

Over $\mathbb{F}_3$ any singular point of $y^2=x^3+ax+b$ would have
$y=0$.  Thus, over $\Q$, any torsion points $(x,y)$ with $3\nmid y$
would be in $E_0(\Q)$, as is the point at infinity; and these would be 
mapped injectively into $E_{{\rm ns}}(\mathbb{F}_p)$.  However
$y^2=x^3+ax+b$ has at most 6 finite points over $\mathbb{F}_3$, since
there are 3 choices for $x$, each corresponding to at most 2 possible
$y$. Thus there are at most 
6 non-trivial torsion points with $3\nmid y$. 

[N; 5 marks]

\newpage

{\bf Solution 3.}

We will show that $q(P+Q)=q(P)q(Q)$ when $P,Q$ lie on $D$ and none of
$P,Q$ or $P+Q$ is $(0,0)$ or the point at infinity. 
Let $P=(u_1,v_1)$, $Q=(u_2,v_2)$ and $P+Q=(u_3,-v_3)$, so that
$(u_1,v_1), (u_2,v_2), (u_3,v_3)$ are collinear, lying on $v=mu+c$,
say, as well as on $E$.  Thus $u_1,u_2,u_3$ the three roots of
$u(u^2+au+b)-(mu+c)^2=0$, so that for the constant term we 
have $c^2=u_1u_2u_3$. None of $u_1,u_2,u_3$ can vanish, since none of
$P,Q, P+Q$ is $(0,0)$. It follows that $q(P)q(Q)=u_1u_2=u_3(c/u_3)^2$,
which is $u_3=q(P+Q)$ in $\Q^*/(\Q^*)^2$.

[B; 6 marks]\bigskip

Let $(u,v)\in D(\Q)$ and let $q(u,v)$ be represented in
$\Q^*/(\Q^*)^2$ by a square-free integer $r$, which could have either
sign. We claim that $r|b_1$, which will show that $q$ has finite
image. The claim is trivial if $u=0$, since $q(u,v)=b_1$ in this case.
Otherwise $r,u$ and $u^2+a_1u+b_1=u(v/u)^2$ are the same modulo squares,
so that we can write $u=rs^2$ and $u^2+a_1u+b_1=rt^2$ for some
$s,t\in\Q$, with $s\not=0$.  Then 
\[r^2s^4+a_1rs^2+b_1=rt^2.\;\;\;(*)\]
To show that $r|b_1$ we
show that there can be no prime $p$ for which $p|r$ but $p\not\mid
b_1$.  If there were such a $p$, there would be integers $e,f$ with
$|r^2s^4|_p=p^{2+4e}$, $|b_1|_p=1$, and $|rt^2|_p=p^{1+2f}$ or 0.
Thus these three terms have different valuations.  Moreover for any
positive real $x\not=1$ we have $\sqrt{x}<\max(x,1)$, whence 
\[|a_1rs^2|_p\le |rs^2|_p<\max(|r^2s^4|_p,1)=\max(|r^2s^4|_p,|b_1|_p.)\]
This provides the required contradiction, since in (*) the two largest
terms would have the same valuation.

[B; 7 marks]\bigskip

We have $A^2+C^2=2B^2$ and $B^2+D^2=2C^2$.  So if $x=-2B^2C^{-2}$ we have
\[x(x^2+5x+4)=x(x+1)(x+4)=(-2B^2C^{-2})(1-2B^2C^{-2})(4-2B^2C^{-2})\]
\[=(-2B^2C^{-2})(-A^2C^{-2})(2D^2C^{-2})=(2ABDC^{-3})^2,\]
giving a point $(x,y)$ with $x=-2B^2C^{-2}$ and $y=2ABDC^{-3}$.

[N; 2 marks]\bigskip

There is a 2-isogeny $\phi:E\rightarrow F$ where $F:V^2=U(U^2+a_1U+b_1)$
with $a_1=-2.5=-10$ and $b_1=5^2-4.4=9$.  The map $q$ above (on $F(\Q)$ 
rather than $D(\Q)$) has kernel $\phi(E(\Q))$.  We claim that the
image is trivial. A coset in the image has a representative $r|b_1=9$ which is
square-free, so that $r=\pm 1$ or $\pm 3$. Moreover (*) is solvable,
leading to
\[rs^4-10s^2+9r^{-1}=t^2.\]
We shall show that only $r=1$ is possible.
If $r<0$ the equation cannot hold, since the LHS is negative.  If $r=3$ we
would get, putting $s=l/m$ in lowest terms, that
$3l^4-10l^2m^2+3m^4=n^2$ 
with integers $l,m,n$ in which $l,m$ are coprime.  
Then $-10(lm)^2\equiv n^2\pmod{3}$, which is possible only if 3
divides both $n$ and $lm$.  Then 9 divides $10(lm)^2$ and $n^2$, whence
$3|l^4+m^4$. This can happen only when both $l$ and $m$ are divisble
by 3, a contradiction to our assumption that $l$ and $m$ were coprime.
Thus the image of $q$ is trivial, and so $\# F(\Q)/\phi(E(\Q))=1$.

[S; 5 marks]\bigskip

There is an analogous homomorphism $\hat{q}$ from $E(\Q)$ to
$\Q^*/(\Q^*)^2$ in which any point in the image has a representative
$r$, which is a square-free integer dividing $b=4$. Thus $r=\pm 1$ or
$\pm 2$, so that the image has size at most 4.  As above we deduce that
$\# E(\Q)/\hat{\phi}(F(\Q))\le 4$.

[S; 1 mark]\bigskip

Since $\hat{\phi}\circ\phi=[2]$ it now follows that
$\# E(\Q)/2E(\Q)\le 4$. However $E(\Q)$ has 4 points of order
dividing 2, so that $E_{{\rm tors}}(\Q)/2E_{{\rm tors}}(\Q)$ has order
4.  The rank is therefore zero.

[S; 2 marks]\bigskip

It follows that the only points are the torsion points, which have
$X=0,-1,-4$ or $-2$.  Thus when $X=-2B^2C^{-2}$ as above we must have
$B=C$ and hence $A,B,C,D$ are all equal. 

[N; 2 marks]



\end{document}
